Theorem equtr 1490

Description: A transitive law for equality.

Assertion

Ref	Expression
equtr	|- (x = y -> (y = z -> x = z))

Proof of Theorem equtr

Step	Hyp	Ref	Expression
1		ax-8 1306	. 2 |- (y = x -> (y = z -> x = z))
2	1	equcoms 1489	1 |- (x = y -> (y = z -> x = z))

Syntax hints:   -> wi 3   = wceq 1298

This theorem is referenced by:  equtrr 1491  equtr2OLD 1493  equequ1 1494  equvin 1652  a12lem1 1767  axsep 3437  dscmet 9196

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 1305  ax-8 1306  ax-12 1310  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481

Copyright terms: Public domain